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Ber [7]
2 years ago
6

The relationship between the decay constant (λ) and the half-life (t1⁄2) is given by the equation t(1/2) = 0.693/λ. Use the equa

tion for N to derive this relationship.
Hint: when t = t(1/2), N/No = 0.5 (1 point)

A sample of carbon-14 initially consists of 5 × 1024 particles. Carbon-14 has a half-life of 5730 years.

a. What is the decay constant for carbon-14? (Answer in units of yr–1.) (1 point)
b. How many radioactive particles of the sample remain after 100 years? (1 point)
c. What percentage of the radioactive particles remains after 500 years? (1 point)
d. How many radioactive particles of the sample remain after 1000 years? (1 point)
e. How much time will it take for 50% of the particles to decay? (1 point)
f. How much time will it take for 99% of the particles to decay? (1 point)
g. How many half-lives will it take for 99% of the sample to decay? (1 point)
h. What is the initial decay rate of the sample? (Answer in decays/yr.) (1 point)
i. After 200 years, what is the decay rate of the sample? (1 point)
j. How long will it take for the decay rate to decrease to 1015 decays/year? (1 point)
k. How many half-lives have passed after the time you found in part (j)? (1 point)
Physics
1 answer:
pshichka [43]2 years ago
6 0

Answer: A: 70/2=35

B: 35/2=17

C: 17.5/2=8.75

D: 8.75 of C-14 will be left

E: 5,730 years

F: 5,631

Explanation:

that’s all I got, hope I helped kinda

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Answer:

13 blocks

Explanation:

The total distance the student travels is 13 blocks.

 Distance is the length of path covered during the motion of a body.

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 Total distance  = 3blocks + 4blocks + 6blocks  = 13blocks

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3 years ago
A packing crate rests on a horizontal surface. It is acted on by three horizontal forces: 600 N to the left, 200 N to the right,
egoroff_w [7]

Answer:

The resultant force would (still) be zero.

Explanation:

Before the 600-N force is removed, the crate is not moving (relative to the surface.) Its velocity would be zero. Since its velocity isn't changing, its acceleration would also be zero.

In effect, the 600-N force to the left and 200-N force to the right combines and acts like a 400-N force to the left.

By Newton's Second Law, the resultant force on the crate would be zero. As a result, friction (the only other horizontal force on the crate) should balance that 400-N force. In this case, the friction should act in the opposite direction with a size of 400 N.

When the 600-N force is removed, there would only be two horizontal forces on the crate: the 200-N force to the right, and friction. The maximum friction possible must be at least 200 N such that the resultant force would still be zero. In this case, the static friction coefficient isn't known. As a result, it won't be possible to find the exact value of the maximum friction on the crate.

However, recall that before the 600-N force is removed, the friction on the crate is 400 N. The normal force on the crate (which is in the vertical direction) did not change. As a result, one can hence be assured that the maximum friction would be at least 400 N. That's sufficient for balancing the 200-N force to the right. Hence, the resultant force on the crate would still be zero, and the crate won't move.

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